How to draw an Archimedean solid that meets certain conditions?

Question

The "Archimedean solid" is also known as a semi-regular solid, which is formed by combining different regular polygons with unequal numbers of sides. As shown in the figure, if we cut off an equilateral pyramid from each of the eight vertices of a cube along the midpoint of each edge, we obtain a semi-regular solid whose surface area is equal to 12 + 4√3. The following statements about this semi-regular solid are correct:

A. AB = √2

B. The surface area of the sphere that circumscribes the semi-regular solid is 6π.

C. The angle between AB and plane BCD is π/4.

D. There are 16 edges that form an angle of π/3 with AB.

Graphics3D[{Opacity[.5], 
  PolyhedronData["Cuboctahedron", "Polyhedron"]}, Boxed -> True, 
 ImageSize -> 300]

How to draw an Archimedean solid that meets the given known conditions as follows?

Answer 1

• The same way as https://mathematica.stackexchange.com/a/289031/72111 and https://mathematica.stackexchange.com/a/289169/72111

Clear["Global`*"];
{pts, lines, center, faces} = 
  PolyhedronData[
   "Cuboctahedron", {"VertexCoordinates", "Lines", "Centroid", 
    "FaceIndices"}];
L = RegionMeasure[lines[[1]]];
λ = Sqrt[2]/L;
pts = λ (# - center) & /@ pts;
polyhedron = Polyhedron[pts, faces]

t = 1.1; 
DynamicModule[{point = 10  {Cos[t], Sin[t], .5}, 
  vertical = {0, 0, 1}, angle = 8  Degree}, 
 Overlay[Graphics3D[#, Lighting -> {{"Ambient", White}}, 
     Boxed -> False, ViewPoint -> Dynamic@point, 
     ViewVertical -> Dynamic@vertical, 
     ViewAngle -> Dynamic@angle] & /@ {{EdgeForm[
      AbsoluteThickness[3]], 
     polyhedron}, {EdgeForm[
      Directive@{AbsoluteThickness[2], 
        AbsoluteDashing[{2, 6}, 0, "Round"]}], FaceForm[], polyhedron}}, All,
   1]]